C Programming - Strings - Discussion

No people the explanation is wrong if the value is call by value, then any change of value t in the function swap should not affect the value of strings pstr[0] and pstr[2] in main.

Try the below code:

#include<stdio.h>
void swap(char *, char *);

int main()
{
char *pstr[2] = {"Hello", "IndiaBIX"};
swap(pstr[0], pstr[1]);
printf("%s\n%s", pstr[0], pstr[1]);
return 0;
}
void swap(char *t1, char *t2)
{
char *t;
t=t1;
t1=t2;
t2=t;
*t1='3';
*t2='4';
}

In Turbo C it works to give output as 4ello 3ndiabix. This proves that the transfer is call by reference only.

The reason there is no actual swapping done in the original code is that we are just swapping the new pointers (t1 and t2) who are pointing to the original strings and not the actual pointers who are pointing to them pstr[0] and pstr[1].

Its like we already had pstr[0] and pstr[1], and now we assigned t1 and t2 respectively and then we discarded t1 and t2 by doing this it will not affect the value of pstr[0] and pstr[2].

Actually we are passing here value of the pointers & catching with single * variable,hence due to call by value its not affecting the original value of variable, in stead of it if we will pass the address of pointer @ catch with ** pointer variable,then due to call by reference the original variable will be affected and swaping will be come in light,by that way swap also done but its not look because change is not done in original variable.
this the one of the way to do swap with affect in original variable.

#include<stdio.h>
void swap(char **, char **);

int main()
{
char *pstr[2] = {"Hello", "IndiaBIX"};
swap(&pstr[0],&pstr[1]);
printf("%s\n%s", pstr[0], pstr[1]);
return 0;
}
void swap(char **t1, char **t2)
{
char *t;
t=*t1;
*t1=*t2;
*t2=t;
}

@All.

According to my knowledge, we can write it as;

#include <stdio.h>
void swap(char **, char **);

int main()
{
char *name = "Utkarsh";
char *title = "Gupta";
swap(&name, &title);
printf("%s\n%s\n", name, title);
return 0;
}

The reason there is no actual swapping done in the original code is that we are just swapping the new pointers (t1 and t2) who are pointing to the original strings and not the actual pointers who are pointing to them pstr[0] and pstr[1].

void swap(char **a, char **b)
{
char *temp;
temp = *a;
*a = *b;
*b = temp;
}

Since,pstr[2] is array of pointers to strings.So two pointer points to two strings.When pstr[0] & pstr[1] are passed ,it passes values pointed to by both pointers.

Run this

#include<stdio.h>
void swap(char *, char *);

int main()
{
char *pstr[2] = {"Hello", "IndiaBIX"};

printf("%s\n%s", pstr[0], pstr[1]);
return 0;
}

It will print the values not an addresses.

How come the following code works as call by reference then?

#include<stdio.h>
void fun(int *p);

int main()
{
int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0};
int *ptr;
ptr = &a[0][0];
fun(ptr);
printf("%d\n", *ptr);
return 0;
}
void fun(int *p)
{
++*p;
printf("%d\n", *p);
}

Some1 please tell me the syntax of calling this swap function by reference so that the string "Hello" and "IndiaBIX" get actually swaped. Me waiting eagerly.

What is call by value and call by reference?

@bhupesh

swap(&pstr[0],&pstr[1]);

Good job @Kamal Nayan.

Thanks @Kamal Nayan.